I've been trying to do this, but I really don't understand how contour integration works. The variable $z=a+ib$ is complex.
\begin{equation}
\oint_C \frac{z\,e^{z\lambda}\,dz}{(z+1)^3}
\end{equation}
For the contour $C$ : $|z-\frac{1}{2}| = 2 $
Do I have to find the poles first, right?
Best Answer
Alternatively, you can use the straight-forward expression for the residue of a function at a pole of order $n$
$${\rm Res}\left(f\left(z\right),z_{0}\right)=\dfrac{1}{\left(n-1\right)!}\dfrac{{\rm d}^{n-1}}{{\rm d}z^{n-1}}\lim_{z\rightarrow z_{0}}\Bigg[\left(z-z_{0}\right)^{n}f\left(z\right)\Bigg]$$
In your case $f\left(z\right)=\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}$, thus $z_{0}=-1$, $n=3$ and so
$${\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\dfrac{1}{2!}\dfrac{{\rm d}^{2}}{{\rm d}z^{2}}\lim_{z\rightarrow -1}\Bigg[\left(z+1\right)^{3}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}\Bigg]=\frac{1}{2}\lambda e^{-\lambda}\left(2-\lambda\right)$$
You then get from the residue theorem that
$$\oint_{C\left(\frac{1}{2},\sqrt{2}\right)}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}{\rm d}z=2\pi i{\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\pi i\lambda e^{-\lambda}\left(2-\lambda\right)$$
exactly as @coreyman317 got.